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Leya [2.2K]
3 years ago
5

What best explains the type of energy present in the vibrating atoms of a substance

Chemistry
1 answer:
luda_lava [24]3 years ago
7 0
The type of energy present in the vibrating atoms of a substance is a thermal energy and it is a kinetic type of energy. It is associated with movement within the crystal lattice of substance. ... Eventually, it can lead to motion of the atoms which is a form of kinetic energy.
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Draw the structure of 4-chlorophthalic acid in the window below.
neonofarm [45]
The structure of <span>4-chlorophthalic acid is attached below:

4-chlorophthalic acid is a dibasic acid i.e. it has 2 replaceable hydrogen atom. As compared to mineral acids, it is a weak acid. However, as compared to aliphatic acid, it is a strong acid. This is because of resonance effect. 
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6 0
3 years ago
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61.0 mol of P4O10 contains how many moles of P
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Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
To make 2.00 L of 0.300 M sulfuric acid, how many mL of a 1.00 M stock solution should be used?
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We are given with
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Does water burn? for my sister that says water does burn
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No, water can’t burn
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